Question

olecular iodine, I2(g), dissociates into iodine atoms at 625 K with a first-order rate constant of 0.271 s-1. (a) What is the half-life for this reaction? s (b) If you start with 0.048 M I2 at this temperature, how much will remain after 5.37 s assuming that the iodine atoms do not recombine to form I2? M

Answer 1

Answer:

* $t_{1/2}=2.56s$

* $[I_2]=0.011M$

Explanation:

Hello,

In this case, considering the reaction:

$I_2(g)\rightarrow 2I$

Which is first-order with respect to I₂, we can compute the half-life by:

$t_{1/2}=\frac{ln(2)}{k}=\frac{ln(2)}{0.271s^{-1}}\\ \\t_{1/2}=2.56s$

Moreover, since the integrated rate law is:

$[I_2]=[I_2]_0exp(-kt)$

We can compute the concentration of iodine once 5.37 s have passed:

$[I_2]=0.048Mexp(-0.271s^{-1}*5.37s)$

$[I_2]=0.011M$

Best regards.